Answer the question 7
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9
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.
Answered by
1
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//
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