Question

(15) determine the correct congruence. which of the following triangle congruences is correct, according to the figure? 1.2.3.4. (א) ( \triangle arg cong \triangle apg ) (ב) ( \triangle arg cong \triangle asp ) (ג) ( \triangle agp cong \triangle asp ) (ד) ( \triangle arp cong \triangle asg )

Explanation:

Step1: Analyze triangle congruence criteria

We need to check which triangles are congruent. Let's look at the given diagram: \( RG = PS \) (marked with one tick), \( AG = AP \) (marked with two ticks), and angles at \( G \) and \( P \) are \( 70^\circ \). Also, \( RP = RG + GP \), \( SG = SP + GP \), so \( RP = SG \) (since \( RG = SP \)).

Step2: Check each option

- Option (א): \( \triangle ARG \cong \triangle APG \). Let's see: \( AG = AG \) (common), \( RG \) and \( PG \) – wait, \( RG \) and \( PG \) are not necessarily equal. Wait, no, maybe I made a mistake. Wait, \( AG = AP \) (given), \( \angle AG R=\angle AP S = 180 - 70=110^\circ \)? Wait, no, the angles at \( G \) and \( P \) are \( 70^\circ \), so \( \angle AGR = 180 - 70 = 110^\circ \), \( \angle APS = 180 - 70 = 110^\circ \). Wait, maybe better to check \( \triangle ARP \) and \( \triangle ASG \). \( RP = SG \) (as \( RG = SP \), so \( RG + GP = SP + GP \)), \( AG = AP \), and \( \angle ARP \) and \( \angle ASG \): since \( AG = AP \), \( \angle APG=\angle AGP = 70^\circ \), so \( \angle ARP = 180 - \angle AGR = 110^\circ \), \( \angle ASG = 180 - \angle APS = 110^\circ \). Wait, no, let's check option (ת): \( \triangle ARP \cong \triangle ASG \). \( RP = SG \) (proven), \( AG = AP \), \( \angle ARP=\angle ASG \) (since \( \angle AGR = \angle APS = 70^\circ \), so supplementary angles \( \angle ARP = \angle ASG = 110^\circ \)? Wait, no, maybe using SAS. \( AR \) and \( AS \): wait, maybe I messed up. Wait, let's check \( \triangle ARP \) and \( \triangle ASG \): \( RP = SG \) (because \( RG = SP \), so \( RG + GP = SP + GP \)), \( AG = AP \) (given), and \( \angle ARP = \angle ASG \) (since \( \angle AGR = \angle APS = 70^\circ \), so \( 180 - 70 = 110^\circ \) for \( \angle ARP \) and \( \angle ASG \)). Wait, but maybe the correct one is \( \triangle ARP \cong \triangle ASG \) (option (ת)). Wait, let's re - evaluate each option:

- Option (א): \( \triangle ARG \cong \triangle APG \): \( AG = AP \), \( \angle AGR=\angle APG = 70^\circ \)? No, \( \angle AGR \) is adjacent to \( 70^\circ \), so \( \angle AGR = 110^\circ \), \( \angle APG = 110^\circ \)? Wait, no, the angle at \( G \) is \( 70^\circ \), so \( \angle AGP = 70^\circ \), \( \angle AGR = 180 - 70 = 110^\circ \), \( \angle APS = 180 - 70 = 110^\circ \). So \( \triangle ARG \) and \( \triangle APG \): sides \( RG \) and \( PG \) – not equal, so not congruent.

- Option (ב): \( \triangle ARG \cong \triangle ASP \): \( AG = AP \), \( RG = SP \) (given), \( \angle AGR=\angle APS = 110^\circ \) (since \( 180 - 70 = 110 \)). So by SAS, \( \triangle ARG \cong \triangle ASP \)? Wait, but \( \angle AGR \) and \( \angle APS \) are \( 110^\circ \), \( RG = SP \), \( AG = AP \). But wait, maybe I made a mistake. Wait, the diagram: \( R - G - P - S \), so \( RG \) and \( SP \) are equal (one tick), \( AG \) and \( AP \) are equal (two ticks). So \( \triangle ARG \) and \( \triangle ASP \): \( RG = SP \), \( AG = AP \), \( \angle AGR=\angle APS = 110^\circ \) (since they are supplementary to \( 70^\circ \)). So SAS would apply. But wait, let's check the correct answer. Wait, maybe the intended answer is \( \triangle ARP \cong \triangle ASG \) (option (ת)). Wait, \( RP = SG \) (as \( RG = SP \), so \( RG + GP = SP + GP \)), \( AG = AP \), and \( \angle ARP = \angle ASG \) (since \( \angle AGR = \angle APS = 70^\circ \), so \( 180 - 70 = 110^\circ \)). So by SAS, \( \triangle ARP \cong \triangle ASG \).

Wait, maybe I confused the angles. Let's start over. The marks: \( RG = SP \) (one tick), \( AG = AP \) (two ticks). So \( RP=RG + GP \), \( SG = SP+GP \), so \( RP = SG \). \( AG = AP \), and \( \angle ARP=\angle ASG \) (because \( \angle AGR=\angle APS = 70^\circ \), so \( 180 - 70 = 110^\circ \) for \( \angle ARP \) and \( \angle ASG \)). So in \( \triangle ARP \) and \( \triangle ASG \): \( RP = SG \), \( \angle ARP=\angle ASG \), \( AP = AG \). So by SAS, \( \triangle ARP\cong\triangle ASG \) (option (ת)).

Answer:

(ת) \( \triangle ARP \cong \triangle ASG \)