The uploaded image describes a geometry problem involving a trapezoid ABCD, with the following information:

• ∠A = ∠D = 90°,
• AB = 10 cm, CD = 30 cm, and AD = 35 cm,
• Point M is on side AD, and AM = 15 cm.

The tasks are:

1. Prove that triangles ABM and DMC are similar.
2. Prove that ∠BMC = 90°.

I will now solve the problem step by step.

Part (a): Prove that △ABM ∼ △DMC

To show that two triangles are similar, we need to prove that they satisfy one of the criteria for similarity (such as AA - Angle-Angle, SAS - Side-Angle-Side, or SSS - Side-Side-Side).

• First, we notice that both triangles share the angle ∠BAM, as both triangles are formed by the same line segments AM.
• Next, since ∠A = ∠D = 90° by given information, we can conclude that ∠ABM = ∠DMC = 90°.

Since we now have two corresponding angles that are equal (∠BAM = ∠DMC and ∠ABM = ∠DMC), by AA similarity criterion, we can conclude that △ABM ∼ △DMC.

Part (b): Prove that ∠BMC = 90°

To prove that ∠BMC = 90°, consider the following:

• Since △ABM ∼ △DMC, the corresponding sides are proportional.
• Now, notice that in both triangles, the angle ∠BAM is common, and both triangles share a right angle at ∠ABM and ∠DMC.
• The sum of angles in any triangle is 180°, so considering the angle sum property and the geometry of the figure, we can deduce that ∠BMC must be a right angle, as it is the complement of ∠BAM and ∠DMC.

Thus, ∠BMC = 90°.

Would you like further details or clarification on any part of this solution?

Related questions:

1. How can you prove similarity in triangles using the SAS criterion?
2. What is the geometric significance of the perpendicular lines in this trapezoid?
3. How does the angle sum property apply to similar triangles?
4. What are other ways to prove two triangles are similar apart from AA?
5. How does proportionality of sides relate to similarity in triangles?

Tip:

In geometry, when working with similar triangles, always check for common angles and proportional sides, as they are often the key to proving similarity.