E is a point on the side ad produced of a parallelogram abcd

Affine maps preserve collinearity, hence we may assume that the original parallelogram $ABCD$ is a square, then embed the construction in $\mathbb\right).$$ The collinearity of $D,P,Q$ is a straightforward consequence of $$ P+\lambda D = (\lambda+1) Q,$$ giving that $Q$ is a convex combination of $P$ and $D$. A simple alternative is just to apply By introducing few extra points, you may also prove the same by symmetry: We can prove the theorem in this equivalent form: if $Q$ is the intersection between $AC$ and $PD$, then $BQ$ is parallel to $DM$. Let $N$ be the intersection between $BQ$ and $CD$, then by similar triangles we have: $$ =AM, $$ where the last equality follows from the intercept theorem applied to parallel lines $PM$ and $AC$. We get then that triangles $CBN$ and $ADM$ are equal. In particular: $\angle CBN=\angle ADM$ so that $BN\parallel DM$.

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:MAT 3300 - Exercises to practice 1. ABCD is a parallelogram. AE is an altitude to side CD. AC is an altitude to side BC, Prove: i) m MAT 3300 - Exercises to practice 1. ABCD is a parallelogram. AE is an altitude to side CD. AC is an altitude to side BC, Prove: i) m

Views: 5,927 a = − 1.5 ; d = − 0.5 purthich of the following situations, the sequence of numbers formed will form an A.P.? (i) The cost of digging a well for the first metre is ₹150 and rises by ₹20 for each [NCERT] (ii) The amount of air present in the cylinder when a vacuum pump removes each time 4 1 ​ of their remaining in the cylinder. [NCERT] (ii) Divya deposited ₹1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year,..., and so on. [NCERT EXEMPLAR] (iii) △ AEP − △ ADB (iv) △ PDC − △ BEC 8. E is a point on the side AD produced of a parallelogram A BC D and BE intersects C D at F Show that △ A BE − △ CFB. 9. In Fig 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: Updated On Jan 21, 2023 Topic All topics Subject Mathematics Class Class 10 Answer Type Video solution: 1 Upvotes 54 Avg. Video Duration 9 min

Transcript Ex 6.3, 8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB Given: A parallelogram ABCD where E is point on side AD produced & BE intersects CD at F To Prove: ΔABE ∼ ΔCFB. Proof: In parallelogram ABCD , opposite angles are equal, Hence, ∠A = ∠C Also, In parallelogram ABCD opposite sides are parallel, AD ∥ BC Now since AE is AD extended, AE ∥ BC and BE is the traversal ∴ ∠ AEB = ∠ CBF Now in Δ ABE & Δ CFB ∠A = ∠C ∠ AEB = ∠ CBF ∴ Δ ABE ∼ Δ CFB Hence proved Show More

Views: 5,598 ∴ a = 9 , d = − 5 , n = 27 , t n = ? ∴ t n = a + ( n − 1 ) d ∴ t n ​ = 9 + ( 27 − 1 ) − 5 ∴ t n = 9 + ( 26 ) − 5 ∴ t n = 9 − 130 ∴ t n = − 121 ​ ∴ The 2 7 th term of the A.P. is − 121. Q.6. Find how many three digit natural numbers are divisible by 5. Sol: List of three digit no. divisible by 5 is. 100 , 105 , 110 , 115 … .995 Views: 5,121 f ( x ) = x 2 + p x + 45 is equal to 144, find the value of p. 16. If α and β are the zeros of the quadratic polynomial f ( x ) = x 2 − p x + q, prove that β 2 α 2 ​ + α 2 β 2 ​ = q 2 p 4 ​ − q 4 p 2 ​ + 2 77. If α and β are the zeros of the quadratic polynomial f ( x ) = a x 2 + b x + c, then evaluate: (i) a α + b 1 ​ + a β + b 1 ​ (ii) a α + b β ​ + a β + b α ​ (iv) a ( β α 2 ​ + α β 2 ​ ) + b ( β α ​ + α β ​ )