Q. 104.3( 6 Votes )
Prove that a circle drawn on one of the equal sides of an isosceles triangle as diameter bisects the base of the triangle.
In the figure ΔABC is an isosceles such that AB = BC. Circle C(O, r) is drawn on the side AB as diameter and intersecting side AC in D.
To Prove: AD = CD
Construction: Join B to D
Proof: ∠ADB = 90° (angle in a semi-circle)
Now in ΔABD and ΔBDC
AB = BC (given)
BD = BD (common)
∠ADB = ∠BDC (each 90°)
∴ ΔABD ≅ ΔBDC
∴ AD = CD
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
RD Sharma | Extra Qs. of Cyclic Quadrilaterals31 mins
Important Questions on Circles29 mins
NCERT | Imp. Qs. on Circles43 mins
Quiz | Lets Roll on a Circle45 mins
Proof of Important Theorems of Circles45 mins
Concept of Cyclic Quadrilateral61 mins
Master Theorems in Circles42 mins
Genius Quiz | Rolling Around Circles49 mins
Proof of Important Theorems of Circles46 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
view all courses
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation