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Question
In Fig. 10.99, AD ⊥ CD and CB ⊥. CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.
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Solution
Given that, in the figure AD ⊥ CD and CB ⊥ CD and AQ = BP,DP =CQ
We have to prove that ∠DAQ=∠CBP
Given that DP= QC
Add PQ on both sides
Given that DP=QC
Add PQ on both sides
⇒ DP+PQ=PQ+QC
⇒ DQ=PC ................(1)
Now, consider triangle DAQ and CBP,
We have
∠ADQ=∠BCP=90° [given]
AQ=BP [given]
And DQ=PC [given]
So, by RHS congruence criterion, we have ΔDAQ≅ΔCBP
Now,
∠DAQ=∠CBP [ ∵Corresponding parts of congruent triangles are equal]
∴ Hence proved
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