QuestionProve that in a given diagram, (i) $A B$ is parallel to $P Q$ and (ii) $M P \times A M = B M \times M Q$ .

Studdy Solution

STEP 1

Assumptions1. $B$ is a tangent to the circle $Q K$ at . $Q$ is a tangent to the circle $K B A$ at $Q$
3. $A M K Q$ is a straight line4. In the circles, lines connecting the center of the circles to the points of tangency are perpendicular to the tangents

STEP 2

(i) To prove $A B$ is parallel to $Q$
We use the property that a radius of a circle is perpendicular to the tangent line at the point of tangency.
Let $O1$ and $O2$ be the centers of the circle $Q K$ and $K B A$ respectively. The segments $O1$ and $O2 Q$ are are radii of their respective circles and are perpendicular to the tangent lines $B$ and $Q$ respectively in accordance with this property.

STEP 3

By using basic geometry, it's known that parallel lines have corresponding angles that are equal.First, let's check the angles $\angle A O2 Q$ and $\angle B O1$ .
Because $O2 Q$ is perpendicular to $Q$ , angle $\angle A O2 Q$ is a right angle.
And since $O1$ is perpendicular to $B$ , angle $\angle B O1$ is also a right angle.So, $\angle A O2 Q = \angle B O1$ since both are90 degrees.

STEP 4

Similarly, we can check the other pair of corresponding angles. Let's examine angles $\angle Q O2 A$ and $\angle O1 B$ now.
Because $O2 A$ and $O1 B$ are extended from the radii $O2 Q$ and $O1$ respectively, the angles $\angle Q O2 A$ and $\angle O1 B$ are vertically opposite and therefore equal. We can write this as $\angle Q O2 A = \angle O1 B$ .

STEP 5

Now that we have both pairs of corresponding angles as equal, we can state that the lines $A B$ and $Q$ are parallel according to the corresponding angles property of parallel lines. Hence, $A B \parallel Q$ .

STEP 6

(ii) To prove $M \times A M = B M \times M Q$
This can be proven by using the properties of similar triangles.Note that in triangle $A M$ and triangle $Q M B$ , the angles $\angle A M$ and $\angle Q M B$ are equal (same reasoning as in step4), and the angles $\angle A M$ and $\angle Q B M$ are equal due to $A B \parallel Q$ established in step5.

STEP 7

Since two of their angles are equal, the triangles $A M$ and $Q M B$ are similar by the AA criterion.

STEP 8

In similar triangles the ratios of their corresponding sides are equal. So we can write $\frac{{A M}}{{ M}} = \frac{{B M}}{{M Q}}$ which implies $M \times A M = B M \times M Q$ That's what we wanted to prove.

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