In figure- ABCD is a parallelogram Prove that DPPQ-DCBQ

∠ DPC=∠ BPQ(Vert. Opp) ∠ DCP=∠ PBQ(Int. Alt) ∠ CDP=∠ BQP(Int. Alt) By AAA criteria #160; Δ PCD∼Δ PBQ Since The triangles are similar th ...

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Standard IX

Mathematics

Opposite Sides of a Parallelogram Are Equal

In figure, ...

Question

In figure, $A B C D$ is a parallelogram
Prove that $\frac{D P}{P Q} = \frac{D C}{B Q}$

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Solution

$∠ D P C = ∠ B P Q (V e r t . O p p) ∠ D C P = ∠ P B Q (I n t . A l t) ∠ C D P = ∠ B Q P (I n t . A l t)$
By $A A A$ criteria
$Δ P C D \sim Δ P B Q$
Since The triangles are similar the corresponding sides proportional .
$⟹ \frac{D P}{P Q} = \frac{D C}{B Q}$

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In figure, ABCD is a parallelogram Prove that DPPQ=DCBQ

∠DPC=∠BPQ(Vert.Opp)∠DCP=∠PBQ(Int.Alt)∠CDP=∠BQP(Int.Alt)By AAA criteria ΔPCD∼ΔPBQSince The triangles are similar the corresponding sides proportional .⟹DPPQ=DCBQ

In figure, $A B C D$ is a parallelogram
Prove that $\frac{D P}{P Q} = \frac{D C}{B Q}$

$∠ D P C = ∠ B P Q (V e r t . O p p) ∠ D C P = ∠ P B Q (I n t . A l t) ∠ C D P = ∠ B Q P (I n t . A l t)$
By $A A A$ criteria
$Δ P C D \sim Δ P B Q$
Since The triangles are similar the corresponding sides proportional .
$⟹ \frac{D P}{P Q} = \frac{D C}{B Q}$