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.

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- Oct 19th 2009, 11:38 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. - Oct 20th 2009, 06:21 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.

- Oct 20th 2009, 11:07 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) - Oct 21st 2009, 04:35 AMmr fantastic
- Oct 27th 2009, 02:21 AMxxravenxx