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In the $\triangle ADC$ , $\angle DAC$ or angle $A$ is a right angle, E is the midpoint of AC . The perpendicular drawn to $AC$ from $E$ meets $DC$ at $B$

i.Drawn the given information in a figure & prove that $\triangle BAD$ is isosceles

ii. $AC^2+AD^2=4AB^2$

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Currently I am unable to think of anything of the problem , Any help would be kindly appreciated , A demonstration proof would help much

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  1. $B$ is the centre of the Thales semicircle, hence $BC$, $BA$, $BD$ are radii and hence equal.
  2. Pythagoras with $CD=2AB$.
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Hint:

Note that the triangle $CBE$ is equal to $ABE$ because they have equal angle in $E$ and equal sides $CE=EA$ and $CE=CE$.

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Clearly $\Delta CEB$ is congruent to $\Delta AEB$ (both right triangles, common legs).

Now we just do some angle chasing. By the congruence, we have $<BCE=<BAC$. Call it $\theta$. Then, letting $\overline \theta=\frac {\pi}2-\theta$ we have $\overline \theta=<CBE=<ABE=<DAB$. It follows that $<DBA=\pi-2\overline \theta$ when in turn implies that $<BDA=\overline \theta$. Thus $\Delta BDA$ is isosceles (two of its angles $=\overline \theta$).

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