In the given right angled triangle ABC, right angled at B. P is the mid-point of AC. Prove that
BP=1/2AC​

The right angled triangle is missing, so i have attached it.

Proved below

Step-by-step explanation:

From the attached image, we can see that QP is parallel to BC.

Now, looking at triangles AQP and ABC, we can see that;

∠ AQP = ∠ ABC

Also;

∠ APQ = ∠ ACB

This means that triangles ABC and AQP are similar.

We are told that P is the mid-point of AC.

Thus; AP = PC

We can now say that;

AP:AC = 1:2 and AQ:AB = 1:2

Since AQ:AB = 1:2, it means Q is the midpoint of AB.

Now, ∠ AQP = ∠ BQP

However, according to SAS congruence test, AQP is congruent to BQP.

Thus; AP = BP = CP

Since AP:AC = 1:2 as we saw earlier, then;

BP:AC = 1:2

Thus:

BP/AC = ½

BP = ½AC

idk

step-by-step explanation:

idk

1: -2 3/4, -2 1/2, -14, 0, 3, 8 1/3

2: -4, -3, -2 1/2, -2 1/4, 3 1/4,

step-by-step explanation: