Question

Answer the question 7​

Answer 1

Answer:

Step-by-step explanation:

consider Δ ADP and Δ CBQ

AD=BC  (opposite sides of a rectangle)

∠DAP=∠BCQ  (AD║BC so are alternate interior angles)

∠DPA=∠BQC = 90°  (DP⊥AC & BQ⊥AC)

hence using AAS similarity

ΔADP≅ΔCBQ

so by CPCT

∠ADP=∠CBQ

DP=BQ.

Answer 2

Answer:

GIVEN:

Abcd is a rectangle

DP⊥AC, BQ⊥AC

TO PROVE:

1)ΔADP≅ΔCBQ

2)∠ADP=∠CBQ

3)DP=BQ

PROOF:

Consider ΔADP & ΔCBQ,

In ΔADP & ΔCBQ,

AD=CB (opposite sides of a rectangle are equal)

A rectangle is a parallelogram, so the oppposite sides will be equal and parallel,

∠DAP=∠BCQ (Alternate angles as AB║CD)

∠DPA=∠BQC ( DP and BQ ⊥ AC)

ΔADP≅ΔCBQ (AAS congruence criterion)

∠ADP=∠CBQ (CPCT)

DP=BQ (CPCT)

hence proved//