Problem 14

Let ABC be a right angled triangle. A circle Г have side AC as its diameter meets hypotenuse AB at point E. A tangent line to Г at point E meets side BC at point D. Prove that a triangle BDE is isosceles.

Printable View

- October 19th 2009, 10:38 PMxxravenxxHelp with this hard geometry problem!
Problem 14

Let ABC be a right angled triangle. A circle Г have side AC as its diameter meets hypotenuse AB at point E. A tangent line to Г at point E meets side BC at point D. Prove that a triangle BDE is isosceles. - October 20th 2009, 05:21 AMSoroban
Hello, xxravenxx!

Quote:

14. Let be a right triangle: .

A circle has side as its diameter, meets hypotenuse at

A tangent line to at meets side at point

Prove that is isosceles.

Code:`F`

o

\

θ'\

A o * \

| θ * *\ E

| θ o

| * * *

| * \ θ ' *

| * *\ *

O * * \ *

| * \ *

| \ *

| * \ *

| * \ *

| * \ θ' *

C o * - - - - - - - o - - - - - - - - - - - - - - o B

D

Draw radius

Since is isosceles.

. .

Since is tangent at

. . Hence, and are complementary.

. . Let

and are vertical angles: .

In and are complementary.

. . Hence: .

In

Therefore, is isosceles.

- October 20th 2009, 10:07 PMxxravenxx
Awesome. Thanks heaps. But a little advice - try to draw a diagram through paint or something instead of typing it up. It makes it a little hard to understand. But I got there.

Thank you. (Rofl) - October 21st 2009, 03:35 AMmr fantastic
- October 27th 2009, 01:21 AMxxravenxx