×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

let ABC be a triangle and D be the midpoint of BC. suppose the angle bisector of angle ADC is tangent to the circumcirle of triangle ABD at D. prove that angle A is equal to 90°. RMO 2016


4 years ago

Shamik Banerjee
27 Points
							Let the bisector of $\angle$ADC be DE where E is a point on the side AC. $\angle$ADE = $\angle$CDE = $\alpha$ (say). As DE is the tangent at D for the circumcircle of $\triangle$ABD, we get $\angle$DAB = $\angle$ADE = $\alpha$. Using exterior angle theorem in $\triangle$ABD, we get $\angle$ADC = $\angle$ABD + $\angle$DAB $\Rightarrow$ 2$\alpha$ = $\alpha$ + $\angle$DAB $\Rightarrow$ $\angle$DAB = $\alpha$. This makes $\triangle$ABD isosceles with BD = AD. So AD = DC as D is the mid-point of BC. Therefore, $\triangle$ADC is isosceles with $AD = AC \Rightarrow \angle DAC = \angle DCA = \frac{180^{\circ} - 2\alpha }{2} = 90^{\circ} - \alpha$ $\angle A = \angle BAC = \angle DAB$$\,$ + $\angle DAC = \alpha$ + $\, 90^{\circ}\, -\, \alpha = 90^{\circ}$

3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Analytical Geometry

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 53 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions