smc-4747-30-AAS

Congruency, SMB-010

In the diagram, `AD` is parallel to `BC`, `AC` bisects `/_BAD` and `BD` bisects `/_ABC`. The lines `AC` and `BD` intersect at `P`.

Prove that `/_BAC = /_BCA`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Prove that `Delta ABP ≡ Delta CBP`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

Show Worked Solution

`text(Prove)\ /_BAC = /_BCA`

`/_BCA`	`= /_CAD\ \ \ text{(alternate angles,}\ BC\ text(\|\|)\ AD text{)}`
`/_CAD`	`= /_BAC\ \ \ text{(}AC\ text(bisects)\ /_BAD text{)}`
`:. /_BAC`	`= /_BCA\ …\ text(as required)`

ii. `text(Prove)\ Delta ABP ≡ Delta CBP`

`/_BAC`	`= /_BCA\ \ \ text{(from part (i))}`
`/_ABP`	`= /_CBP\ text{(}BD\ text(bisects)\ /_ABC text{)}`
`BP\ text(is common)`

`:. Delta ABP ≡ Delta CBP\ \ text{(AAS)}`

In the figure below, \(ABCD\) is a parallelogram where opposite sides of the quadrilateral are equal.

Prove that a diagonal of the parallelogram produces two triangles that are congruent. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\text{One of multiple solutions:}\)

\( AB=CD\ \ \text{and}\ \ AC=BD\ \ \text{(given)} \)

\(BC\ \text{is common} \)

\(\therefore\ \Delta ABC \equiv \Delta DCB\ \ \text{(SSS)}\)

In the figure below, \(AB \parallel DE, \ AC = CE\) and the line \(AE\) intersects \(DB\) at \(C\).

Prove that this pair of triangles are congruent. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\( \angle ACB = \angle DCE\ \ \text{(vertically opposite)} \)

\(AC = CE\ \ \text{(given)} \)

\(\angle BAC = \angle DEC\ \ \text{(alternate,}\ AB \parallel DE \text{)} \)

\(\therefore\ \Delta ABC \equiv \Delta EDC\ \ \text{(AAS)}\)

In the quadrilateral \(ABCD\), \(AB \parallel CD, \angle BAD = \angle BCD\) and \(\angle DBC = \angle BDA = 90^{\circ} \).

Prove that this pair of triangles are congruent. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\angle BAD = \angle BCD\ \ \text{(given)} \)

\(\angle DBC = \angle BDA = 90^{\circ} \ \ \text{(given)} \)

\(BD\ \text{is common} \)

\(\therefore\ \Delta BAD \equiv \Delta DCB\ \ \text{(AAS)}\)