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Title:

Grade: 10-a Lesson: S2-L1

Explanation:

From the figure we can observe that

AB is a linesegment, m is a line, P is a point on line m and m \$\bot\$ AB.

$2$

m \$\bot\$ AB and passes through C which is the mid-point of AB.

We have to show that \$PA = PB\$.

Consider

\begin{align} \triangle PCA and \triangle PCB\\ \end{align}

We have AC = BC since C is the mid-point of AB

\begin{align} \angle PCA = \angle PCB = 90° \tag{Given}\\ PC = PC \tag{common} \\ \end{align}

So,

\begin{align} \triangle PCA \cong \triangle PCB \tag{SAS rule}\\ \end{align}

\$∴ PA = PB\$ as they are corresponding sides of congruent triangles.

From the figure we can observe that

$2$

Steps	Statment	Solution
1	Given	From the figure we can observe that AB is a linesegment, m is a line, P is a point on line m and m \$\bot\$ AB.
2		m \$\bot\$ AB and passes through C which is the mid-point of AB.
3	Side of triangle	We show that PA = PB
4	Angles	\$\triangle PCA and \triangle PCB\$
5		We have AC = BC since C is the mid-point of AB
6	Given	\$\angle PCA = \angle PCB = 90°\$
7	Common	PC = PC
6	Congruency	\$\trianglePCA \cong \trianglePCB\$
7	Since	\$∴ PA = PB\$ as they are corresponding sides of congruent triangles.

Steps

Statment

Solution

Given

From the figure we can observe that
AB is a linesegment, m is a line, P is a point on line m and m \$\bot\$ AB.

m \$\bot\$ AB and passes through C which is the mid-point of AB.

Side of triangle

We show that PA = PB

Angles

\$\triangle PCA and \triangle PCB\$

We have AC = BC since C is the mid-point of AB

Given

\$\angle PCA = \angle PCB = 90°\$

Common

PC = PC

Congruency

\$\trianglePCA \cong \trianglePCB\$

Since

\$∴ PA = PB\$ as they are corresponding sides of congruent triangles.