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The internal bisector of angle A of triangle ABC meets the circumcircle in D. If DE and DF are the perpendiculars to AB and AC respectively from D. Prove that AE is arithematic mean of AB and AC .

Rishabh Shetty
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2 Answers2

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Let $ABCD$ be as in the diagram below.

enter image description here

Then $AB=2R\cos\theta$ and $AC=2R\cos\phi$.

So $$\frac{AB+AC}{2}=R\cos\theta+R\cos\phi=2R\cos(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})=AD\cos(\frac{\theta-\phi}{2})$$ which is the perpendicular distance. (Note that angle $DAC=(\theta-\phi)/2$.)

Chrystomath
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  • Thanks for answering my question Chrystomath and Matteo . However , I had a hard time understanding trignometry and found Matteo ' s proof easy to understand . BTW , Matteo asked me some question in his post 1) Because of LP and opposite angles of cyclic quad. 2)csct . Thanks again both, but how did you came to know that we how to proof triangle CDF and BDE congruent ? – Rishabh Shetty Mar 21 '19 at 03:36
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Of course no trigonometry is required. Considering the Figure below, we need to prove that $\triangle CDF \cong \triangle BDE$. This is straightforward by noting the following.

  1. $DE \cong DF$, since $AD$ is bisector of $\angle CBD$.
  2. $\angle DFA \cong \angle DEB$, being both right angles, by construction.
  3. $\angle DBE \cong \angle DCF$, since they're both supplementary w.r.t $\angle ACD$ (can you tell why?)

The thesis follows from the fact that $CF \cong EB$. Can you tell why?

enter image description here

dfnu
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    What's wrong with trigonometry? You don't need E and F! – Chrystomath Mar 20 '19 at 14:40
  • @Chrystomath nothing wrong. As a personal approach, I always stick with the bare necessities. And since trigonometry is just a direct consequence of computing ratios, I use those directly. Also, in my country, Euclidean geometry is taught much earlier than trigonometry. Consequently, I always tend to avoid using it when helping out students. – dfnu Mar 20 '19 at 15:32
  • Thanks for answering my question Chrystomath and Matteo . However , I had a hard time understanding trignometry and found Matteo ' s proof easy to understand . BTW , Matteo asked me some question in his post 1) Because of LP and opposite angles of cyclic quad. 2)csct . Thanks again both, but how did you came to know that we how to proof triangle CDF and BDE congruent ? – – Rishabh Shetty Mar 21 '19 at 03:36
  • @RishabhShetty I manipulated the AM in order to get something that could be easilier translated into a congruence between segments that appeared in the original construction (as a first try I generally avoid to draw more lines than the ones implied by the original text. Also when I see "circle" I immediatly reason in terms of relationship between angles and/or distances from the center (not in this case).

    In order to avoid downvotes and help the community helping you, next time I suggest you post also your doubts and difficulties in solving the problem.

     – dfnu Mar 21 '19 at 08:16