Ex.10.5 Q11 Circles Solution - NCERT Maths Class 9

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Question

\({ABC}\) and \({ADC}\) are two right triangles with common hypotenuse \({AC.}\) Prove that \(\angle{CAD}= \angle {CBD}.\)

 Video Solution
Circles
Ex 10.5 | Question 11

Text Solution

What is given ?

\({ABC}\) and \({ADC}\) are two right triangles with common hypotenuse \({AC.}\)

What is unknown?

Proof of \(\angle{CAD}= \angle {CBD}\)

Reasoning:

Sum of all angles in a triangle is \(180^\circ\)

If the sum of pair of opposite angles in a quadrilateral is \(180\) then it is cyclic quarilateral

Angles in the same segment of a circle are equal.

Steps:

Consider \(\Delta {ABC,}\)

\(\begin{align}&\angle {ABC}+  \angle {BCA}+\angle {CAB} =180^{\circ} \\&\text { (Angle sum property of a triangle) }\end{align}\)

\(\begin{align} 90^{\circ}+\angle {BCA}+\angle {CAB}&=180^{\circ} \\ \angle {BCA}+\angle {CAB}&=90^{\circ} \ldots(1)\end{align}\)

Consider \(\Delta {ADC,}\)

\(\begin{align}&\angle {CDA}+  \angle {ACD}+\angle {DAC} =180^{\circ} \\&\text { (Angle sum property of a triangle) }\end{align}\)

\(\begin {align} 90^{\circ}+\angle {ACD}+\angle {DAC}&=180^{\circ}\\\angle {ACD}+\angle {DAC}&=90^{\circ} \ldots(2) \end {align}\)

Adding Equations (\(1\)) and (\(2\)), we obtain

\( \begin {align} \begin{Bmatrix} \angle {BCA}+\angle {CAB}\\+\angle {ACD}+\angle {DAC} \end{Bmatrix}&=180^{\circ} \\ \begin{Bmatrix} (\angle {BCA}+\angle {ACD})+ \\ (\angle {CAB}+\angle {DAC}) \end{Bmatrix} &=180^{\circ} \end {align}\)

\( \begin {align}  \angle {BCD}+\angle {DAB}&=180^{\circ} \ldots(3) \end {align}\)

However, it is given that

\(\begin {align}\angle {B}+\angle {D} &=90^{\circ}+90^{\circ} \\ &=180^{\circ} \ldots(4) \end {align}\)

From Equations (\(3\)) and (\(4\)), it can be observed that the sum of the measures of opposite angles of quadrilateral \(\text{ABCD}\) is \(180^{\circ}.\) Therefore, it is a cyclic quadrilateral.

Consider chord \({CD.}\)

\(\begin {align}\angle {CAD} = \angle {CBD} \end {align}\) (Angles in the same segment)

  
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