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

the highest percentage for deaths in construction from highest to lowest is as follows: falls, struck by objects, electrocution, and caught-in/between. these four are collectively known as fatal four.

in the case of falls, the type that is responsible for the overall highest death in the category is falls from roofs.

2x + 8

step-by-step explanation: