Q7 of 16 Page 133

In the given figure AD is perpendicular to CD and BC is perpendicular to CD. If AQ = BP and DP=CQ, Show that ∠DAQ=∠CBP.


Given: AD ⊥ CD and BC ⊥ CD


AQ = BP and DP = CQ


To Prove: ∠DAQ = ∠CBP


Proof:


AD ⊥ CD and BC ⊥ CD


∴ ∠D = ∠C (∵ each angle is 90°)


∵ DP = CQ (Given)


Adding PQ to both sides. We get


DP + PQ = PQ + CQ


⇒ DQ = CP …(i)


Now, in right angles ADQ and BPC, we have


∴ AQ = BP (given)


Also, DQ = CP (from equation (i))


Therefore, we can say that


ΔADQ ≅ ΔBPC by Right angle hypotenuse side (RHS) rule of congruency.


∴ ∠DAQ = ∠CBP (Since Corresponding part of congruent triangles are always equal)


Hence, proved.


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