Step-4

Title: Congruency of triangles(SAS)

Grade: 10-a Lesson: S2-L5

Explanation:

Description: 4

d4

Steps Statment Solution

1

Given

AC is altitude of △ABC, AC ⊥ BD and C is the mid-point of BD, side ∠ABC = 10M − 1 and ∠ADC = 170 − M.

2

Consider \$\triangle ACB\$ and \$\triangle ACD\$

\begin{align} EC &= EC \\ \angle ACB &= \angle ACD \\ BC &= DC \\ \therefore \triangle ACB & \cong \triangle ACD \text{(by SAS congruency)} \end{align}

3

From \$\triangle ACB\$ and \$\triangle ACD\$

\begin{align} \angle CBA &= \angle CDA \\ 10M - 1 &= 170 - 9M \\ 19M &= 171 \\ M &= 19 \\ \end{align}

d4

Given,

AC is altitude of △ABC, AC ⊥ BD and C is the mid-point of BD,

sides ∠ABC = 10M − 1 and ∠ADC = 170 − M.

Altitude is the height of the triangle, it bisects the base of the triangle perpendicularly.

Consider \$\triangle ACB\$ and \$\triangle ACD\$

\begin{align} EC &= EC \tag{Common side} \\ \angle ACB &= \angle ACD \tag{AC ⊥ BD} \\ BC &= DC \tag{C is the mid-point of BD} \\ \therefore \triangle ACB & \cong \triangle ACD \text{(by SAS congruency)} \\ \end{align}

From \$\triangle ACB\$ and \$\triangle ACD\$

\begin{align} \angle CBA &= \angle CDA \\ 10M - 1 &= 170 - 9M \\ 19M &= 171 \\ M &= 19 \\ \end{align}

\begin{align} \text{∴ The value of M is 19. } \end{align}


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