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  • Please solve RD Sharma class 12 chapter Scaler and Dot Product exercise 23.2 question 9 maths textbook solution

Please solve RD Sharma class 12 chapter Scaler and Dot Product exercise 23.2 question 9 maths textbook solution

Answers (1)

Answer: \angle A D C=\angle A D B=90^{\circ}

Hint: Firstly prove the two triangle form by median

Given: if the median to base of triangle is perpendicular to base of triangle is isosceles

Solution:

 Let ABC an isosceles triangle with AB=AC and let AB be median

To base BC then D Is midpoint of BC

\angle A B C=\angle A C B\\\\\ in\; \triangle \mathrm{ABD} \; \& \; \triangle \mathrm{ACD}\\\\ A B=A C \quad (equal \; side )

\begin{aligned} &\angle A B C=\angle A C B \quad \text { (equal angle }) \\\\ &B D=D C \quad(\mathrm{BC} \text { bisected by median } \mathrm{AD}) \end{aligned}

Therefore by SAS criterion of congurence triangle

\begin{aligned} &\triangle \mathrm{ABD} \cong \triangle \mathrm{ACD} \\\\ &\text { Hence } \angle A B C=\angle A C B \end{aligned}        (by correcponding part of congurence)

Let \; \angle A D C=\angle A D B=x\\\\ \angle A D B+\angle A D C=180^{\circ}

Put the value we get

\begin{aligned} &x+x=180^{\circ} \\\\ &2 x=180^{\circ} \\\\ &x=90^{\circ} \\\\ &\angle A D C=\angle A D B=90^{\circ} \end{aligned}

 

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