Question

A triangle ABC has a ratio of AB : ( AB + BC ) = 5 : 8 such that BD bisects an angle ABC and M is the midpoint between BD. If the area of ABC is 240 units^2, then what is the area of triangle ABM?

AGAIN NO SILLY ANSWER PLZ

Answer 1

Step-by-step explanation:

In ΔABC, we have

∠B=2∠C or, ∠B=2y, where ∠C=y

AD is the bisector of ∠BAC. So, let ∠BAD=∠CAD=x

Let BP be the bisector of ∠ABC. Join PD.

In ΔBPC, we have

∠CBP=∠BCP=y⇒BP=PC

In Δ

′

s ABP and DCP, we have

∠ABP=∠DCP, we have

∠ABP=∠DCP=y

AB=DC [Given]

and, BP=PC [As proved above]

So, by SAS congruence criterion, we obtain

ΔABP≅ΔDCP

⇒∠BAP=∠CDP and AP=DP

⇒∠CDP=2x and ∠ADP=DAP=x [∴∠A=2x]

In ΔABD, we have

∠ADC=∠ABD+∠BAD⇒x+2x=2y+x⇒x=y

In ΔABC, we have

∠A+∠B+∠C=180

∘

⇒2x+2y+y=180

∘

⇒5x=180

∘

[∵x=y]

⇒x=36

∘

Hence, ∠BAC=2x=72

Answer 2

Answer:

I think 93.75 unit^2

if it is correct... reply me... i'll give step by step explanation