Question

In triangle ABC, AD is a median. FE is a straight line parallel to BC cutting the remaining sides AB at F and AC at E and cutting the median at X.

If ED bisects angle ADC:


Triangle AFX is similar to Triangle ABD and that Triangle AEX is similar to Triangle ACD.


Using above information prove that angle FDE=90

Answer 1

Answer:

Step-by-step explanation:

since AD is a median it implies that triangle ABC is bisected to two equal right angled triangle which are ADB and ADC.

FE is parrallel to BC and cuts AB at F and AC at E shows that there are two similar triangles formed which are AFE and ABC.

Recall that ADC is a right angled triangle, ED bisects a right angled triangle the the ADE = $45^{o}$.

Now, Let FD bisect angle ADB,

  then ADF = $45^{o}$ too.

Since AFX is similar to Triangle ABD and that Triangle AEX is similar to Triangle ACD, then EDX is similar to FDX

FDE = ADF + ADE = $90^{o}$