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$AD \bot CD$ and $CB \bot CD$. If AQ=BP and DP=CQ , prove that $\angle DAQ = \angle CBP$
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Last updated date: 20th Jun 2024
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Hint: Given that, $AD \bot CD$ and $CB \bot CD$. Now we have to prove $\angle DAQ = \angle CBP$. Note that both $\vartriangle ADQ$ and $\vartriangle BPC$ are right angled triangles. AQ=BP and DP=CQ, given. Therefore, first we have to show that the triangles are congruent. And lastly, show that $\angle DAQ = \angle CBP$, as they are corresponding parts of congruent triangles.

Complete step-by-step solution:
Given, $AD \bot CD$ and $CB \bot CD$.
$ \Rightarrow \angle ADQ = \angle BCP = {90^ \circ }$
Therefore, both $\vartriangle ADQ$ and $\vartriangle BPC$ are right angled triangles.
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Also, AQ=BP and DP=CQ
$ \Rightarrow DP + PQ = CQ + PQ$
$ \Rightarrow DQ = CP$
Now, in $\vartriangle ADQ$ and $\vartriangle BPC$,
$\angle ADQ = \angle BCP = {90^ \circ }$
AQ=BP (given)
Therefore, $\vartriangle ADQ \cong \vartriangle BPC$ (by RHS rule of congruence)
Hence, $\angle DAQ = \angle CBP$ (corresponding parts of congruent triangles)

Note: The four rules of congruency are as follows:
SSS: When three sides of two different triangles are equal in length.
SAS: When two sides are equal, and the angle between them is also the same in measure.
AAS: When any two angles and a side is equal.
RHS: When the hypotenuse and any one side of two right angled triangles are equal in length.