**Question:**

ABCD is a rectangle. M and N are points on AB and DC respectively. MC and BN meet at X. M is the midpoint of AB.

(a) Prove that and are similar.

(b) Given that area of : area of =9:4, find the ratio of,

(i) DN: NC

(ii) area of rectangle ABCD: area of . **(Challenging)**

**[Answer Key]** (b) (i) 1:3

(ii) 20:3

**Suggested Solutions:**

(a)

(vert. opp. angles)

(alt. angles)

(alt. angles)

Therefore, and are similar (AAA).

(b) (i)

Let and

Then

So

Thus,

(ii)

**We now have a shorter solution, thanks to a visitor to our site! (see comments below)**

From part (a), since and are similar, we have

This means that

Thus (the two triangles share a common height)

Now, note that

Hence area of

We conclude that area of rectangle ABCD: area of

Here is a longer solution, for those who are interested:

Let area of

Let area of

Let area of

We have since and have the same base BC and their heights have ratio 3:2.

Cross-multiplying, we get

So

since and have the same base BC and their heights have ratio 3:4.

Hence,

Thus, area of

area of rectangle ABCD: area of =40:6=20:3

alternative methode to b(ii) :

\triangle CXN and \triangle MXB are similar, so MX / CX = 2 / 3, implies that MC / CX = 5 / 3. Thus \triangle MBC : \triangle XBC = 5 / 3

since Area of rectangle ABCD = 4 * \triangle MBC (M is midpoint of AB), so

area of rectangle ABCD: area of \triangle XBC = (4*5) : 3 = 20 : 3

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Thanks!

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It is easier to see

Rectangular ABCD is 4x MBC

if we draw a dotted line down from midpoint M to divide the rectangula into 2 halves.

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