Question

point a, b, c are on a circle, such that m (arc BC)=m(arc AB)=120° no point, except point b, is common to the arcs .what is the type of triangle ABC

Answer 1

In triangle ABC, M and N are two points on BC such that the angle BAM is equal to angle BCA. How do you prove that AB square is equal to BC into MB?



In △ABC,M△ABC,M is a point on BCBC such that ∠BAM=∠BCA.∠BAM=∠BCA.

We have to prove that AB2=BC×MB.AB2=BC×MB.



Whenever we get such a problem, convert what is to be proved into ratios.

So, we have to prove that ABBC=MBAB.ABBC=MBAB.

We can now try to see if we can get two similar triangles whose ratio of sides correspond to what is required to be proved.

The numerators are ABAB and MB.MB.So, try to get a triangle having these two as sides.

The denominators are BCBC and AB.AB. So, try to get a triangle having these two as sides.

In △ABM△ABM and △ABC,△ABC,

∠BAM=∠BCA∠BAM=∠BCA (given)

and ∠ABM∠ABM is common.

⇒△ABC∼△ABC⇒△ABC∼△ABC (A-A test).

⇒ABBC=MBAB⇒ABBC=MBAB (c.s.s.t.)

⇒AB2=BC×MB.

Biitch, please.

Answer 2

Answer:

equilateral triangle