The diagonals of a quadrilateral abcd intersect each other at the point o such that show that abcd is a trapezium.

Given: The diagonals of a quadrilateral ABCD intersect each other at the point O such that B O A O ​ = D O C O ​ i.e., C O A O ​ = D O B O ​ To Prove: A B C D is a trapezium Construction: Draw O E ∥ D C such that E lies on B C . Proof: In △ B D C, By Basic Proportionality Theorem, O D B O ​ = E C B E ​ . . . . . . . . . . . . ( 1 ) But, C O A O ​ = D O B O ​ (Given) . . . . . . . . . ( 2 ) ∴ From ( 1 ) and ( 2 ) C O A O ​ = E C B E ​ Hence, By Converse of Basic Proportionality Theorem, O E ∥ A B Now Since, A B ∥ O E ∥ D C ∴ A B ∥ D C Hence, A B C D is a trapezium.

The diagonals of a quadrilateral ABCD intersect each other at the point Osuch that AO/BO = CO/DO. Show that ABCD is a trapezium Solution: We know according to the In Draw OE || AB In ΔABC, OE || AB ⇒ OA/OC = BE/CE (Basic Proportionality Theorem) ---------- (1) But, OA/OB = OC/OD (given) ⇒ OA/OC = OB/OD ----------- (2) From equations (1) and (2) OB/OD = BE/CE In ΔBCD, OB/OD = BE/CE OE || CD (Converse of Basic proportionality theorem) We know that, OE || AB and OE || CD Thus, AB || CD Hence we can say ABCD is a ☛ Check: Video Solution: The diagonals of a quadrilateral ABCD intersect each other at the point Osuch that AO/BO = CO/DO. Show that ABCD is a trapezium Summary: The diagonals of a quadrilateral ABCD intersect each other at the point such that AO/BO = CO/DO. Hence it is proved that ABCD is a trapezium. ☛ Related Questions: • • • •

Hint: We will start by drawing the figure of quadrilateral \[ABCD\] such that \[\dfrac \\ \] So, we have that in \[\Delta ACD\], the sides \[AC\], \[AD\] are divided in equal ratios. Thus, we can apply the converse of Basic proportionality theorem to get, side \[CD\] parallel to line \[EO\]. So, we have \[ AB||EO||CD \\ \Rightarrow AB||CD \\ \] We know that if two sides in a quadrilateral are parallel to each other, then the quadrilateral is called a trapezium. Hence proved that \[ABCD\] is trapezium. Note: The condition required to prove that a quadrilateral is a trapezium is to show that two opposite sides of the quadrilateral are parallel. Use the Basic proportionality theorem that states that, if we draw a line parallel to a side of the triangle so that it intersects with the other two sides, then the other two sides are divided in equal ratios. We have used this theorem because this theorem deals with the property of parallel lines in a triangle. And a quadrilateral can be easily divided in triangles.

Let us consider the following figure for the given question. Draw a line OE || AB In ΔABD, OE || AB By using basic proportionality theorem, we obtain `(AE)/(ED) = (BO)/(OD) ....(1)` However, it is given that `(AO)/(OC) = (OB)/(OD) ....(2)` From equation 1 nd 2 we obtain `(AE)/(ED) = (AO)/(OC)` ⇒ EO || DC [By the converse of basic proportionality theorem] ⇒ AB || OE || DC ⇒ AB || CD ∴ ABCD is a trapezium. • 3 more Video Tutorials For All Subjects • Similarity of Triangles video tutorial 00:16:50 • Similarity of Triangles video tutorial 00:09:47 • Similarity of Triangles video tutorial 00:27:04 • Similarity of Triangles video tutorial 00:25:01 • Similarity of Triangles video tutorial 00:10:21 • Similarity of Triangles video tutorial 00:11:05 • Similarity of Triangles video tutorial 00:27:15

Step 1. Explaining the diagram. Let A B C D be quadrilateral where A C and B D intersects each other at O such that, A O B O = C O D O Step 2. Showing A B C D is trapezium Construction-From the point O , draw a line E O touching A D at E in such a way that, E O ∥ D C ∥ A B I n Δ D A B , E O | | A B By using Basic Proportionality Theorem D E E A = D O O B . . . . . . . . . . . . . . . . . . . . . . . . ( i ) Also, given, A O B O = C O D O ⇒ A O C O = B O D O [ a p p l y i n g a l t e r n e n d o ] ⇒ C O A O = D O O B [ [ a p p l y i n g i n v e r t e n d o ] ⇒ D O O B = C O A O . . . . . . . . . . . . . . . . . . . . . . . . . . ( i i ) From equation ( i ) a n d ( i i ) , We have D E E A = C O A O Therefore, By applying converse of Basic Proportionality Theorem, E O | | D C Also E O | | A B ⇒ A B | | D C . Hence, quadrilateral A B C D is a trapezium with A B | | C D .

As a minor suggestion, I think it is clearer to mark the diagram with information we know will be true (subject to our subsequent proofs). In this case, when writing the proofs, there is a stronger visual connection between the diagram and what is being written. The way it is done in the video, each time an angle is referred to in the proof, I find myself looking at the diagram and following the 3 letters to see the angle, as opposed to sighting a symbol already marked on the diagram identifying the angle. Lets say the two sides with just the < on it where extended indefinitely and the diagonal he is working on is also extended indefinitely just so you can see how they are alternate interior angles. if two lines are both intersect both a third line, so lets say the two lines are LINE A and LINE B, the third line is LINE C. the intersection of LINE A with LINE C creates 4 angles around the intersection, the same is also true about the LINE B and LINE C. There is a quadrant/direction for each of the 4 corners of the angles. So there would be angles of matching corners for each of the two intersections. Now alternate means the opposite of the matching corner. So it's one angle from one intersection and the opposite corner angle from the matching corner on the other intersection. So we have a parallelogram right over here. And what I want to prove is that its diagonals bisect each other. So the first thing that we can think about-- these aren't just diagonals. These are lines ...

Const : Draw a line OM || AB. Proof: In ∆ADB, we have OM || AB. Therefore, by using Basic proportionality theorem, we have [Taking reciprocals of both sides] It is given that, ...(ii) Comparing (i) and (ii), we get Therefore, by using converse of basic proportionality theorem, we have OM || DC But OM || AB (by construction) ⇒ AB || DC Hence, ABCD is a trapezium. Given: AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. To Prove: ∠A > ∠C and ∠B > ∠D Construction: Join AC Proof: In ∆ABC, AB AB ∴ ∠BAC > ∠BCA ...(1) | Angle opposite to longer side is greater In ∆ACD, CD > AD | ∴ CD is the longest side of quadrilateral ABCD ∴ ∠CAD > ∠ACD ...(2) | Angle opposite to longer side is greater From (1) and (2), we obtain ∠BAC + ∠CAD > ∠BCA + ∠ACD ⇒ ∠A > ∠C Similarly, joining B to D, we can prove that ∠B > ∠D.

Since, the shape is not given, let us consider the following cases of a quadrilateral with coordinates of four corners given : Case 1 : Nonuniform quadrlateral. Knowing the coordinates of ABCD, we can find the equation of AC & BD using the standard for of equation, #(y-y1) / (y2 - y1) = (x-x1) / (x2-x1)# By solving the two equations we can find the intersection point O. Case 2 : Square or rectangle or rhombus or parallelogram In all these four shapes, diagonals bisect each other. Using mid point formula, we can find the intersection point. #(x,y) = ((x1 + x2) /2, (y1 + y2) / 2)# Case 3 : Trapezoid or Kite Similar to a quadrilateral. By solving the equations of the diagonals, one can arrive at the intersection point.