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I spend so much time for proving this triangle and i still don't know.

Question :

Given Triangle ABC, AD and BE are altitudes of the triangle. Prove that Triangle DEC similarity with triangle ABC

Annita
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    We normally recommend that you show some of your work in your question, otherwise the question will remain unanswered. It is very good that you have made effort, but attempt to put whatever you have done so far in the question, so that we can get a measure of how far you have reached so far. This will help us serve you better. – Sarvesh Ravichandran Iyer Nov 03 '16 at 12:38
  • Could you prove that $AB$ is parallel with $DE$ . – hamam_Abdallah Nov 03 '16 at 12:40
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    @AbdallahHammam In fact $AB$ and $DE$ are not parallels. The quadrilateral $ABDE$ is cyclic, so it can be proven that $\angle CAB =\angle CDE$. – Ángel Mario Gallegos Nov 03 '16 at 12:43
  • @астонвіллаолофмэллбэрг Okay, thanks a lot for the advice! but, it was my first time to asking, i can't post a picture yet. Next time i'll attach some pictures about my work. Thanks alot! – Annita Nov 03 '16 at 14:56
  • @Annita You are welcome to this site. – Sarvesh Ravichandran Iyer Nov 03 '16 at 23:43

2 Answers2

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Since $AD=AC\cdot cos C$ and $BE=BC\cdot cos C$ we have that $$\frac{AD}{BE}=\frac{AC}{BE}$$ And since the angle that form those two pairs are equal ($\angle C$) the result follows

arberavdullahu
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Construct a figure from the data you have given, now follow these steps(you just have to prove equality of any two angles of respective triangles):

Since BE and AD are perpendiculars so you get ∠BEA = ∠ADB =90 Degrees.Now you can see that ABDE is a cyclic quadrilateral {as ∠BEA = ∠ADB(you can call them angles in the same segment)}

Now as you know sum of opposite angles in a cyclic quadrilateral is 180 degrees so ∠EDB= 180-∠A = ∠EDC. Therefore:

For triangles CDE and CAB: ∠CDE=∠A and ∠C = ∠C

Hence,triangles CDE and CAB are similar.

Vidyanshu Mishra
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  • How did you know if ABDE is a cyclic quadrilateral? Is that because we don't know if DE and AB are parallel or not? – Annita Nov 03 '16 at 14:58
  • U must have heard of angle in the same segment in a circle are equal,the fact i m using here is that converse off the theorem is also true(i.e. if angles in the same segment are equal then quadrilateral is cyclic. – Vidyanshu Mishra Nov 03 '16 at 15:01
  • As i have mentioned that ∠BEA = ∠ADB thus u can think of these angles making a pair of angles in same segment of a circle and since these are equal so quadrilateral is cyclic. – Vidyanshu Mishra Nov 03 '16 at 15:03
  • Okay, i understand it now. So if there are angles in the same segment that equal, then it must be cyclic quadrilateral. Thanks a lot! – Annita Nov 03 '16 at 15:22
  • Sorry for asking again. What if CF is also an altitude of Triangle ABC, can we prove AD is bisector of FDE angle? – Annita Nov 03 '16 at 16:10