Question

in a triangle ABC,BD is a median, if BD=8 cm and BC=14cm, triangle ADB=70 degree and triangle ACB=40 degree. find the area of triangle ABC.​

Answer 1

Answer:

Here is a solution without any trigonometry or calculations.

Let AHAH be the altitude of triangle ΔABCΔABC through vertex AA, i.e. the segment AHAH is perpendicular to ACAC with HH on ACAC. The angle ∠BCH=∠BCA=30∘∠BCH=∠BCA=30∘ in the right-angled triangle BCHBCH yields that ∠CBH=60∘∠CBH=60∘ . Furthermore, DD is the midpoint of segment BCBC, so HDHD is the median of the right angled triangle ΔBCHΔBCH which allows us to conclude that triangle BDHBDH is equilateral and that BD=CD=HD=HBBD=CD=HD=HB. Moreover, ∠CHD=30∠CHD=30.



Angle chasing in triangle ΔACDΔACD shows that ∠HAD=∠CAD=15∘∠HAD=∠CAD=15∘. But together with that,

∠HDA=∠HDB−∠ADB=60∘−45∘=15∘